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fgiga [73]
3 years ago
5

(y4 – y3 + 2y2 + y – 1) ÷ (y3 + 1)

Mathematics
2 answers:
marta [7]3 years ago
7 0

Answer:

The correct answer is 2y^2

Step-by-step explanation:

Monica [59]3 years ago
5 0
<span>a+b) • (a2-ab+b2)
</span><span>(a+b) • (a2-ab+b2) = 
a3-a2b+ab2+ba2-b2a+b3 =
a3+(a2b-ba2)+(ab2-b2a)+b3=
a3+0+0+b3=
a3+<span>b<span>3
ANWSER </span></span></span><span> y4 - y3 + 2y2 + y - 1 —————————————————————— (y + 1) • (y2 - y + 1)</span>
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Suppose it is known that 60% of radio listeners at a particular college are smokers. A sample of 500 students from the college i
vladimir1956 [14]

Answer:

The probability that at least 280 of these students are smokers is 0.9664.

Step-by-step explanation:

Let the random variable <em>X</em> be defined as the number of students at a particular college who are smokers

The random variable <em>X</em> follows a Binomial distribution with parameters n = 500 and p = 0.60.

But the sample selected is too large and the probability of success is close to 0.50.

So a Normal approximation to binomial can be applied to approximate the distribution of X if the following conditions are satisfied:

1. np ≥ 10

2. n(1 - p) ≥ 10

Check the conditions as follows:

 np=500\times 0.60=300>10\\n(1-p)=500\times(1-0.60)=200>10

Thus, a Normal approximation to binomial can be applied.

So,  

X\sim N(\mu=600, \sigma=\sqrt{120})

Compute the probability that at least 280 of these students are smokers as follows:

Apply continuity correction:

P (X ≥ 280) = P (X > 280 + 0.50)

                   = P (X > 280.50)

                   =P(\frac{X-\mu}{\sigma}>\frac{280-300}{\sqrt{120}}\\=P(Z>-1.83)\\=P(Z

*Use a <em>z</em>-table for the probability.

Thus, the probability that at least 280 of these students are smokers is 0.9664.

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